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For a National Board Exam Review:

Find the equation of the ellipse having a length of latus rectum of ${ \frac{3}{2} }$ and the distance between the foci is ${ 2\sqrt{13} }$

Answer is ${ \frac{x^2}{16} + \frac{y^2}{3} = 1 }$

So I try:

$${ LR = \frac{2b^2}{a} = \frac{3}{2} }$$

$${ a^2 - b^2 = c^2 }$$ $${ a^2 - b^2 = ( 2\sqrt{13} )^2 }$$

Solve two equations; I get a = 7.59...

From there I'm stuck. I cant use the variable for the answer... What is wrong with my method?

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    $\begingroup$ I would say $a^2 - b^2 = (2\sqrt{13}/2 )^2$ $\endgroup$ – georg Aug 20 '15 at 14:15
  • $\begingroup$ why divide it by 2? is it not distance between two foci? should I consider it between vertex? @georg $\endgroup$ – james Aug 20 '15 at 14:18
  • $\begingroup$ It is the distance betwen the foci and the center of the ellipse --> the distance between the foci is divided by two. $\endgroup$ – georg Aug 20 '15 at 14:29
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$c=13$, $b^2=\frac{3a}4$

Substituting in $a^2-b^2=c^2$,

$$a^2-\frac{3a}{4}=13$$

Multiply by $4$ to cancel fraction and make it whole.

$$4a^2-3a-52=0$$

$a=4$ and $a=\frac{-13}{4}$

Since we need positive use $4$ to substitute in $$\tag{1}b^2=\frac{3a}4$$

$b^2=3$ and substitute the value of $b$ in $(1)$. You can have $a^2=16$.

Therefore; $$\frac{x^2}{16}+\frac{y^2}{3}=1$$ or $$3x^2+16y^2=48$$

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